是否可以证明 L 是正则语言？
[英]Is it possible to prove that L is a regular language?
问题描述
Let L = {a^f(m) | m >= 1 }
令L = {a^f(m) | m >= 1 } L = {a^f(m) | m >= 1 } where f: Z^+ -> Z^+ is monotone increasing and complies that for all element n in Z^+ there is an m belonging to Z^+ such that f(m+1) - f(m) >= n . L = {a^f(m) | m >= 1 }其中f: Z^+ -> Z^+是单调递增的并且符合对于Z^+中的所有元素n有一个属于Z^+的m使得f(m+1) - f(m) >= n 。
Is it possible to prove that L is a regular language?是否可以证明 L 是正则语言？
1 个解决方案
解决方案1
1 2011-08-16 19:10:21
Let f(x) = 2^x.令 f(x) = 2^x。 For any positive n, f(n+1) - f(n) >= n.对于任何正数 n，f(n+1) - f(n) >= n。
L = {a^f(m)} is not regular. L = {a^f(m)} 不规则。 Consider the strings a^(2^x + 1).考虑字符串 a^(2^x + 1)。 After an FA processes such a string, the smallest string which leads to an accepting state is a^(2^x - 1), having length 2^x - 1. Therefore, a separate state will be needed for every value of x.在 FA 处理这样的字符串之后，导致接受 state 的最小字符串是 a^(2^x - 1)，长度为 2^x - 1。因此，对于每个 x 值都需要单独的 state。 Since there are infinitely many values of x (positive integers), no FA exists to recognize L;由于 x（正整数）的值有无穷多个，因此不存在识别 L 的 FA； ergo, L is not a regular language.因此，L 不是常规语言。
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